Suppose there is a large reservoir of water, and a small horizontal pipe at one of the walls of the pipe. I have done some calculations using Bernoulli's theorem to show that the velocity in the pipe when the end is opened is approximately $$v(t)=2k\tanh\left(\frac{kt}{L}\right)$$ for some positive constant $k$. (It is approximate since we assume the height of the reservoir stays the same, and the water near the top is not affected).

Question: Calculate the timescale for the flow to reach its maximum speed.

I don't understand how to do this, since $\tanh$ doesn't ever reach its maximum. However since they say "timescale" rather than "time", perhaps this means something slightly different? If someone could help me to understand this, and give some idea on how to do this, it would be appreciated.

  • $\begingroup$ When $kt/L$ reaches roughly $2$ (an $O(1)$ constant), the $\tanh$ will be nearly 1, so I'd say the timescale is $L/k$. $\endgroup$ – David May 12 '17 at 0:17
  • $\begingroup$ @David Would you say $L/k$, or $L/kt$? Is a general rule for finding a timescale of functions like this (exponentials too) to take $(\text{coefficient of t})^{-1}$? $\endgroup$ – John Doe May 12 '17 at 12:38
  • $\begingroup$ I'm not sure of any general rules, but my idea was that to make an $O(1)$ change in $\tanh$, you need an $O(1)$ change in $kt/L$, so $t$ has to change by something that is $O(L/t)$. Similar arguments should work in many situations. $\endgroup$ – David May 12 '17 at 16:12
  • $\begingroup$ @David Oh ok, thanks :) $\endgroup$ – John Doe May 12 '17 at 17:57

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